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I have heard that wind affects the actual airspeed that I should fly for maximum range in an aircraft.

I understand that wind will not affect the airspeed that I should fly if I am looking for maximum endurance, but what about maximum range?

How (and how much) should airspeed be adjusted for wind in order to get maximum range?

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  • $\begingroup$ uniform wind (always constant head or tailwind) wouldn't affect best airspeed I believe. $\endgroup$ – ratchet freak Apr 28 '14 at 13:50
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    $\begingroup$ @ratchetfreak From my understanding, it does. You should fly faster into a headwind in order to minimize the time that it affects you, and vice versa for tailwinds. $\endgroup$ – Lnafziger Apr 28 '14 at 13:53
  • $\begingroup$ @Lnafziger - That's why he said uniform winds not variable winds :). But I would imagine there is some strategy to finding tail winds and avoiding headwinds. Much like navigating a river...you know,finding a strong current going the same way you are. $\endgroup$ – Jay Carr Apr 28 '14 at 16:34
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    $\begingroup$ @JayCarr I am referring to uniform winds as well. A uniform headwind or a uniform tailwind. :) $\endgroup$ – Lnafziger Apr 28 '14 at 16:36
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    $\begingroup$ @JayCarr No, my question is regarding a particular wind condition, not one that is changing. $\endgroup$ – Lnafziger Apr 28 '14 at 17:02
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My answer was rather concise first, and I got the impression that I need to elaborate the question first. The question is about the best airspeed for maximum range. With wind. Best range means you cover the most distance while the wind is carrying the plane with it. If you have a headwind, the longer you stay aloft, the more you are carried back, so you better hurry up. With a tailwind, it helps to slow down because now the wind is helping you to cover even more distance.

But how much? We need to pick that particular speed where the change in fuel consumption just balances the change in speed over ground. I always found that easier to explain with gliders, and there you can really observe which polar point is best. Just picture yourself as an observer on the ground who sees the plane fly by in the distance. If you plot a line with the combination of positions and altitude, there must be one flight speed that produces a line where the flight path angle is the shallowest. This is the desired optimum. That has only little to do with optimum L/D - this is just one other point you can find with a sink speed polar. And it happens to be the point for best glide in still air. But there is so much more that humble polar will tell you, if you look at it the right way.

With powered aircraft you need to pick the polar point where your fuel flow is lowest for the given speed over ground. Basically, you fly like the glider and add enough power to stay at the same altitude. That is all the difference. User2168 has answered that part already with a graphical solution.

OK, now back to gliders. Please look at the plot below which shows airspeed on the X axis and sink speed on the Y axis. Optimum Speed for Best Range

The solution is graphical: You start on the X-axis at the point which corresponds to the wind speed and put a tangent on the sink speed graph. Where the tangent touches the sink polar (blue line), the plane flies at the best L/D for that given wind speed. Move the starting point to positive speeds for headwind and to negative speeds (not shown here) for tailwind. If the term "best L/D" is already reserved in your mind, please read this as the "best polar point". It is really the same.

Since User2168 has beaten me to the graphical solution, I will add an analytical solution.

For powered flight things become more tricky, because thrust changes with speed. To simplify things, we can say that thrust changes over speed in proportion to the expression $v^{n_v}$ where $n_v$ is a constant which depends on engine type. Piston aircraft have constant power output, and thrust is inverse with speed over the speed range of acceptable propeller efficiencies, hence $n_v$ becomes -1 for piston aircraft. Turboprops make some use of ram pressure, so they profit a little from flying faster, but not much. Their $n_v$ is -0.8 to -0.6. Turbofans are better in utilizing ram pressure, and their $n_v$ is -0.5 to -0.2. The higher the bypass ratio, the more negative their $n_v$ becomes. Jets (think J-79 or even the old Jumo-004) have constant thrust over speed, at least in subsonic flow. Their $n_v$ is approximately 0. Positive values of $n_v$ can be found with ramjets - they develop more thrust the faster they move through the air.

Now for fuel flow: This goes up and down with the power output of the engine. Again a simplification, but it helps to get to grips with the problem and gives useful results. This lets us re-formulate the problem as: At what airspeed do I have the best ratio between power and ground speed?

Mathematically, we want to fly with $\frac{P}{v_w+v}$ at the lowest possible value. $P$ is the power, $v_w$ is the wind speed and $v$ the airspeed. To express the thrust behavior over speed, I break P up into a product of a constant $K_S$, the throttle setting $\delta$ and the speed like this: $K_S\cdot\delta\cdot v\cdot v^{n_v}$. Here is the general solution, pasted as a PNG to save me all the typing in the editor: formulas

Please note that implicitly the lift coefficient is on both sides of the equation. To solve it, you need to do it recursively, until speed and lift coefficient match. I took this form because of the similarity to the general form at still wind which can be found in many performance books. This here I didn't find anywhere, and it took me a while to figure it out. Thank you, Lnafziger, for the excellent question! It gave me a chance to learn something.

Now I have put the results into a plot. In order to eliminate the aircraft-specific parameters, it shows the ratio of $c_L$ with wind over $c_L$ without wind. The plot is metric, but will work for all units if you use the same units for wind speed and air speed. best cl over speed

To give an example for the application of the correction factor: If you are flying in a 20 m/s headwind and your best range speed at still wind is 50 m/s (approx 97 kts), the $c_L$ needs to be 70% of the $c_L$ at still wind for piston-powered aircraft. This makes your corrected airspeed 60 m/s (v is proportional to $\sqrt{c_L}$), and now the recursive nature of the formula rears its ugly head. At 60 m/s, the correction is only 77.5%, so we need to do a few loops until we arrive at a point where the airspeed and the correction factor match. In this example, this would be 57 m/s or 110 kts in case of a piston-powered aircraft.

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  • $\begingroup$ I have a similar chart in my answer for a powered aircraft. However, I'm not convinced that the best L/D should change with a headwind. I am convinced that the speed that gives best range changes, but airspeed that gives the best lift-to-drag ratio should not be affected by wind. Comments? $\endgroup$ – user2168 Apr 29 '14 at 1:28
  • $\begingroup$ So the "best L/D ratio" means the L/D ratio that provides the best glide path over the ground? I thought best L/D ratio meant largest L/D ratio, which isn't affected by the wind. Is this wrong? If you are asked "what is the best L/D ratio", are you saying you can't answer without knowing the wind? $\endgroup$ – user2168 Apr 29 '14 at 15:28
  • $\begingroup$ @user2168: What is best depends on the circumstances. The question makes it clear that the polar point for best range with wind is requested. And that does change with wind. $\endgroup$ – Peter Kämpf Apr 29 '14 at 19:59
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    $\begingroup$ I still believe that you mean to say glide ratio instead of best L/D here. Glide ratio is in reference to the ground, while L/D is simply lift -vs- drag and has nothing to do with ground track. $\endgroup$ – Lnafziger Apr 29 '14 at 22:21
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    $\begingroup$ The formula contains aircraft-specific values, so there is no way to make generalized, specific conclusions. I would suggest to compile a table of optimum speeds for different wind speeds, wing loadings and altitudes for a specific aircraft. And please keep in mind that this formula uses the parabolic polar approximation. If you know your airplane's polar in more detail, the quality of the table can be improved further. $\endgroup$ – Peter Kämpf May 2 '14 at 22:45
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Consider the case where you are flying into a headwind that equals the published speed for maximum range in still air.

In this case, if you fly at the speed for maximum range in still air, you will have a ground speed of zero, and a range of zero.

To make any progress over the ground (to increase your range), you'll need to fly at an airspeed higher than the speed for maximum range in still air.

The actual relationship is shown in Figure 5-19 of The Illustrated Guide to Aerodynamics (2nd Edition). Plot the Velocity-PowerRequired curve. Draw a line L starting at the point (velocity=headwind, power=0) that is tangent to the Velocity-PowerRequired curve. The point at which the line L touches the PowerRequired curve is at the speed for maximum range.

enter image description here

I think the chart caption is wrong. The chart shows the difference in speeds for maximum range, but I don't believe that the airspeed that gives the best L/D ratio is affected by wind.

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  • $\begingroup$ @JanHudec Do you have a reference for "The maximum L/D is at the minimum of power". My understanding was that the maximum L/D coincided with the tangent when the line in the diagram is drawn from the origin. $\endgroup$ – user2168 Apr 30 '14 at 17:23
  • $\begingroup$ I am sure it could be dug up from How It Flies. But it should be obvious. In steady flight, lift is constant (exactly to balance weight) and power is equal to drag (not accelerating). To get max L/D given constant L we are looking for lowest D, which is lowest power. $\endgroup$ – Jan Hudec Apr 30 '14 at 17:28
  • $\begingroup$ If you had fixed angle of attack, L would change with (square of) speed. But in steady flight, lift balances the weight and angle of attack is adjusted to achieve that balance. Since the graph is for steady flight, it is for constant lift (equal to weight, that's why there are separate graphs by weight). $\endgroup$ – Jan Hudec Apr 30 '14 at 17:33
  • $\begingroup$ You are right; best L/D occurs on the tangent, because the graph shows power, but $D$ in $L/D$ is a force and that puts it on the tangent (from origin). $\endgroup$ – Jan Hudec Apr 30 '14 at 18:52
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Airspeed is your speed through an air mass, wind speed or direction will not change your airspeed. What wind direction changes is your ground speed, which is the speed of the aircraft over the surface of the earth. A headwind component reduces your ground speed, while a tailwind increases it. What I think you are asking is how you change your airspeed to best use fuel depending on whether you have a headwind or tailwind.

The general answer to this is that if you have a headwind you would increase your airspeed, and with a tailwind reduce it. If you have a headwind you will take longer to get where you are going, so increasing airspeed will get you there faster and you will use less fuel. A tailwind gets you there faster so you can reduce your airspeed and therefore fuel burn and still get there in the same amount of time as you would if you were in still air.

In practice whether you do this depends on how much of a headwind/tailwind component you have, and what your burn rates are at different airspeeds. Aircraft have a speed/drag curve, at some point the energy you add to go faster just ends up being wasted. You can also get behind the drag curve, flying too slow to be efficient. Adding a bit more speed may make sense, adding a load more speed will likely end up in using more fuel without much result. The opposite is also true, you can end up getting too slow and draggy, burning more fuel.

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    $\begingroup$ So how do we determine how much to correct? Can you provide an example? $\endgroup$ – Lnafziger Apr 28 '14 at 19:08
  • $\begingroup$ In my case I fly light aircraft, Cessna 172s and PA-28s. If I had a substantial headwind I would increase from 2300 rpm to 2450 or so rpm to add another 5-10 knots of cruise - any more isn't good for the engine. If I had a big tailwind I might reduce to 2100 rpm and lose 5-10 knots of cruise. It's not much for me really, airliners have much more speed to play with, they have software to figure that out. $\endgroup$ – GdD Apr 28 '14 at 19:21
  • $\begingroup$ With a tailwind I believe you would get max range by flying at max endurance airspeed. It gets a bit more complicated with a head wind as the answers above indicate. $\endgroup$ – Michael Hall Oct 5 '18 at 0:07
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This simply boils down to energy required to move from point A to point B, conveniently measured in fuel consumption, or, with gliders, distance gained per altitude given. Under no wind conditions, under power we fly at the air speed where wing and engine/prop are most efficient, near Vy, and with gliding, Vbg.

Going into the wind is where you pay the devil his dues. This becomes critical when you have an emergency power out into a 20 knot headwind and your landing field is ahead of you. Might be easier to turn and fly to a field you already passed (altitude permitting). Another alternative is increasing speed above Vbg to get more distance into the wind. It's no longer about the airspeed, it's about respect to ground. You must reach that point anyway you can to land safely.

A chart or graph of max glide distance W.R.T. ground into headwinds would be extremely useful information for any aircraft (listed as best airspeed under those conditions).

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